Question concerning: $P(n)=\left\lfloor \sqrt{n!+{n^2(n+1)\over 2}}\right\rfloor, n\ge2$

Question

Let $$P(n)=\left\lfloor \sqrt{n!+{n^2(n+1)\over 2}}\right\rfloor, n\ge2$$

$\lfloor x \rfloor$ is the floor function.

Where $P_k$ and $P_j$ are two highest prime numbers such that,

$$P_j \le P(n^2-1)$$

$$P_k \le P(n^2)$$

Then we have $$P_k-nP_j=c$$ The value of $c$ is very small even the values of $P_j$ and $P_k$ are very large.

For examples:

$P(3)$ then $3\le 4$

$P(4)$ then $7\le 8$

$$7-2\times3=1\tag1$$

$P(8)$ then $199\le 201 $

$P(9)$ then $601\le 602 $

$$601-3\times 199=4\tag2$$

$P(15)$ then $1143529\le 1143535$

$P(16)$ then $4574123\le 4574143$

$$4574123-4\times1143529=7\tag3$$

Can $c$ be represent in term of a function of $n$, $$c=f(n)?$$

If not exact... can it $c\approx f(n)?$

Answer 1

According to PARI/GP, the first $25$ values are :

? p
%40 = (n)->sqrtint(n!+n^2*(n+1)/2)
? s
%41 = (m)->precprime(p(m))
? vector(25,q,s(q^2)-q*s(q^2-1))
%42 = [0, 1, 4, 7, 2, 361, 90, 715, 1228, 867, 7320, -149, 1800, 12627, 1456, 35
17, 9090, 6389, 18186, 23113, 9550, 74739, 27120, 19301, 132154]
?

$p$ and $s$ are the necessary self-defined functions.

Some more values :

? q=26;print(q,"  ",s(q^2)-q*s(q^2-1))
26  58797
? q=27;print(q,"  ",s(q^2)-q*s(q^2-1))
27  56476
? q=28;print(q,"  ",s(q^2)-q*s(q^2-1))
28  36835
? q=29;print(q,"  ",s(q^2)-q*s(q^2-1))
29  145260
? q=30;print(q,"  ",s(q^2)-q*s(q^2-1))
30  15049